Introductory maths analysis chapter 03 official

INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences

2007 Pearson Education Asia

Chapter 3 Chapter 3 Lines, Parabolas and Systems Lines, Parabolas and Systems


INTRODUCTORY MATHEMATICAL ANALYSIS

0. Review of Algebra

1. Applications and More Algebra

2. Functions and Graphs

3. Lines, Parabolas, and Systems

4. Exponential and Logarithmic Functions

5. Mathematics of Finance

6. Matrix Algebra

7. Linear Programming

8. Introduction to Probability and Statistics


9. Additional Topics in Probability

10. Limits and Continuity

11. Differentiation

12. Additional Differentiation Topics

13. Curve Sketching

14. Integration

15. Methods and Applications of Integration

16. Continuous Random Variables

17. Multivariable Calculus

INTRODUCTORY MATHEMATICAL ANALYSIS


• To develop the notion of slope and different forms of equations of lines.

• To develop the notion of demand and supply curves and to introduce linear functions.

• To sketch parabolas arising from quadratic functions.

• To solve systems of linear equations in both two and three variables by using the technique of elimination by addition or by substitution.

• To use substitution to solve nonlinear systems.

• To solve systems describing equilibrium and break-even points.

Chapter 3: Lines, Parabolas and Systems

Chapter ObjectivesChapter Objectives


Lines

Applications and Linear Functions

Quadratic Functions

Systems of Linear Equations

Nonlinear Systems

Applications of Systems of Equations

3.1)

3.2)

3.3)

3.4)

3.5)


Chapter OutlineChapter Outline

3.6)


Slope of a Line

• The slope of the line is for two different points (x1, y1) and (x2, y2) is


3.1 Line3.1 Line

change horizontal

change vertical

12

12

xx

yym


The line in the figure shows the relationship between the price p of a widget (in dollars) and the quantity q of widgets (in thousands) that consumers will buy at that price. Find and interpret the slope.


3.1 Lines

Example 1 – Price-Quantity Relationship


Solution:

The slope is


3.1 Lines

Example 1 – Price-Quantity Relationship

2

1

28

41

12

12

qq

ppm

Equation of line

• A point-slope form of an equation of the line through (x1, y1) with slope m is

1212

12

12

xxmyy

mxx

yy


Find an equation of the line passing through (−3, 8) and (4, −2).

Solution:

The line has slope

Using a point-slope form with (−3, 8) gives


3.1 Lines

Example 3 – Determining a Line from Two Points

7

10

34

82

m

026710

3010567

37

108

yx

xy

xy



3.1 Lines

Example 5 – Find the Slope and y-intercept of a Line

• The slope-intercept form of an equation of the line with slope m and y-intercept b is . cmxy

Find the slope and y-intercept of the line with equation y = 5(3-2x).

Solution:

Rewrite the equation as

The slope is −10 and the y-intercept is 15.

1510

1015

235

xy

xy

xy


a.Find a general linear form of the line whose slope-intercept form is

Solution:

By clearing the fractions, we have


3.1 Lines

Example 7 – Converting Forms of Equations of Lines

43

2 xy

01232

043

2

yx

yx


b. Find the slope-intercept form of the line having a general linear form

Solution:

We solve the given equation for y,


3.1 Lines

Example 7 – Converting Forms of Equations of Lines

0243 yx

2

1

4

3

234

0243

xy

xy

yx



3.1 Lines

Parallel and Perpendicular Lines

• Parallel Lines are two lines that have the same slope.

• Perpendicular Lines are two lines with slopes m1 and m2 perpendicular to each other only if

21

1

mm



3.1 Lines

Example 9 – Parallel and Perpendicular Lines

The figure shows two lines passing through (3, −2). One is parallel to the line y = 3x + 1, and the other is perpendicular to it. Find the equations of these lines.



3.1 Lines

Example 9 – Parallel and Perpendicular Lines

Solution:

The line through (3, −2) that is parallel to y = 3x + 1 also has slope 3.

For the line perpendicular to y = 3x + 1,

113

932

332

xy

xy

xy

13

1

13

12

33

12

xy

xy

xy



3.2 Applications and Linear Functions3.2 Applications and Linear FunctionsExample 1 – Production Levels

Suppose that a manufacturer uses 100 lb of material to produce products A and B, which require 4 lb and 2 lb of material per unit, respectively.

Solution: If x and y denote the number of units produced of A and B, respectively,

Solving for y gives

0, where10024 yxyx

502 xy



3.2 Applications and Linear Functions

Demand and Supply Curves

• Demand and supply curves have the following trends:




Example 3 – Graphing Linear Functions

Linear Functions

• A function f is a linear function which can be written as 0 where abaxxf

Graph and .

Solution:

12 xxf 3

215 ttg




Example 5 – Determining a Linear Function

If y = f(x) is a linear function such that f(−2) = 6 and f(1) = −3, find f(x).

Solution: The slope is .

Using a point-slope form:

321

63

12

12

xx

yym

xxf

xy

xy

xxmyy

3

3

236 11



3.3 Quadratic Functions3.3 Quadratic Functions

Example 1 – Graphing a Quadratic Function

Graph the quadratic function .

Solution: The vertex is .

• Quadratic function is written aswhere a, b and c are constants and

22 bxaxxf0a

1242 xxxf

212

4

2

a

b

2 and 6

260

1240 2

x

xx

xx

The points are



3.3 Quadratic Functions

Example 3 – Graphing a Quadratic Function

Graph the quadratic function .Solution:

762 xxxg

312

6

2

a

b

23 x



3.3 Quadratic Functions

Example 5 – Finding and Graphing an Inverse

From determine the inverse function for a = 2, b = 2, and c = 3.

Solution:

cbxaxxfy 2



3.4 Systems of Linear Equations3.4 Systems of Linear Equations

Two-Variable Systems

• There are three different linear systems:

• Two methods to solve simultaneous equations:

a) elimination by addition

b) elimination by substitution

Linear system(one solution)

Linear system(no solution)

Linear system(many solutions)



3.4 Systems of Linear Equations

Example 1 – Elimination-by-Addition Method

Use elimination by addition to solve the system.

Solution: Make the y-component the same.

Adding the two equations, we get . Use to find

Thus,

323

1343

xy

yx

12128

39129

yx

yx

3x

1

391239

y

y

1

3

y

x

3x




Example 3 – A Linear System with Infinitely Many Solutions

Solve

Solution: Make the x-component the same.

Adding the two equations, we get .

The complete solution is

12

5

2

125

yx

yx

25

25

yx

yx

00

ry

rx

52




Example 5 – Solving a Three-Variable Linear System

Solve

Solution: By substitution, we get

Since y = -5 + z, we can find z = 3 and y = -2. Thus,

63

122

32

zyx

zyx

zyx

63

5

1573

zyx

zy

zy

1

2

3

x

y

z




Example 7 – Two-Parameter Family of Solutions

Solve the system

Solution:

Multiply the 2nd equation by 1/2 and add to the 1st equation,

Setting y = r and z = s, the solutions are

8242

42

zyx

zyx

00

42 zyx

sz

ry

srx

24



3.5 Nonlinear Systems3.5 Nonlinear Systems

Example 1 – Solving a Nonlinear System

• A system of equations with at least one nonlinear equation is called a nonlinear system.

Solve (1)

(2)

Solution: Substitute Eq (2) into (1),

013

0722

yx

yxx

7 or 8

2 or 3

023

06

071322

2

yy

xx

xx

xx

xxx



3.6 Applications of Systems of Equations3.6 Applications of Systems of Equations

Equilibrium

• The point of equilibrium is where demand and supply curves intersect.



3.6 Applications of Systems of Equations

Example 1 – Tax Effect on Equilibrium

Let be the supply equation for a manufacturer’s product, and suppose the demand equation is .

a. If a tax of $1.50 per unit is to be imposed on the manufacturer, how will the original equilibrium price be affected if the demand remains the same?

b. Determine the total revenue obtained by the manufacturer at the equilibrium point both before and after the tax.

50100

8 qp

65100

7 qp





Solution:

a. By substitution,

and

After new tax,

and

100

50100

865

100

7

q

qq 5850100100

8p

70.5850.51100100

8p

90

65100

750.51100

100

8

q

q





Solution:

b. Total revenue given by

After tax, 580010058 pqyTR

52839070.58 pqyTR




Break-Even Points

• Profit (or loss) = total revenue(TR) – total cost(TC)

• Total cost = variable cost + fixed cost

• The break-even point is where TR = TC.

FCVCTC yyy




Example 3 – Break-Even Point, Profit, and Loss

A manufacturer sells a product at $8 per unit, selling all that is produced. Fixed cost is $5000 and variable cost per unit is 22/9 (dollars).

a. Find the total output and revenue at the break-even

point.

b. Find the profit when 1800 units are produced.

c. Find the loss when 450 units are produced.

d. Find the output required to obtain a profit of $10,000.




Example 3 – Break-Even Point, Profit, and Loss

Solution:

a. We have

At break-even point,

and

b.

The profit is $5000.

50009

22

8

qyyy

qy

FCVCTC

TR

900

50009

228

q

qq

yy TCTR

72009008 TRy

5000500018009

2218008

TCTR yy

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